Yes, there is -- but this is not really about determinants. Let me introduce
some notations and state the main result.
Definition. Let $\mathbb{N}$ be the set $\left\{ 0,1,2,\ldots
\right\} $. For each $i\in\mathbb{N}$ and $j\in\mathbb{N}$, we let $\left[
i,j\right] $ be the interval $\left\{ i,i+1,\ldots,j\right\} $ of
$\mathbb{N}$ (this is empty when $i>j$).
For each $n\in\mathbb{N}$, we let $\left[ n\right] ^{\prime}$ be the
interval $\left[ 0,n-1\right] =\left\{ 0,1,\ldots,n-1\right\} $ of
$\mathbb{N}$.
Definition. Let $n\in\mathbb{N}$, and let $\sigma$ be a permutation
of $\left[ n\right] ^{\prime}$. Then, $\sigma$ is said to be nimble if for
each $i\in\left\{ 0,1,\ldots,n-1\right\} $, the number $i+\sigma\left(
i\right) +1$ is a power of $2$.
Yes, $1$ counts as a power of $2$. The name "nimble" hints at the game of Nim
(and Nim addition), but I don't have the time to figure out the exact connection.
Here is my main claim (which you conjectured):
Theorem 1. Let $n\in\mathbb{N}$.
(a) Then, there is a unique nimble permutation $\sigma_{n}$ of
$\left[ n\right] ^{\prime}$.
(b) This permutation $\sigma_{n}$ has sign $\left( -1\right)
^{\sigma_{n}}=\left( -1\right) ^{n\left( n-1\right) /2}$.
This permutation yields your conjecture that $\det\left( H_{n}\right)
=\left( -1\right) ^{n\left( n-1\right) /2}$ for each $n\in\mathbb{N}$.
Indeed, if we fix $n\in\mathbb{N}$, then the determinant of the matrix
$H_{n}=\left( a_{i+j}\right) _{i\in\left[ n\right] ^{\prime};\ j\in\left[
n\right] ^{\prime}}$ rewrites as
\begin{align*}
\det\left( H_{n}\right) & =\sum_{\sigma\text{ is a permutation of }\left[
n\right] ^{\prime}}\left( -1\right) ^{\sigma}\underbrace{a_{0+\sigma\left(
0\right) }a_{1+\sigma\left( 1\right) }\cdots a_{n-1+\sigma\left(
n-1\right) }}_{\substack{=
\begin{cases}
1, & \text{if }\sigma\text{ is nimble};\\
0, & \text{if }\sigma\text{ is not nimble}
\end{cases}
\\\text{(by the definition of the }a_{k}\text{)}}}\\
& =\sum_{\sigma\text{ is a permutation of }\left[ n\right] ^{\prime}
}\left( -1\right) ^{\sigma}
\begin{cases}
1, & \text{if }\sigma\text{ is nimble};\\
0, & \text{if }\sigma\text{ is not nimble}
\end{cases}
\\
& =
\sum_{\sigma\text{ is a nimble permutation of }\left[ n\right] ^{\prime}
}\left( -1\right) ^{\sigma} \\
& =\left( -1\right) ^{n\left( n-1\right) /2}\qquad\left( \text{by
Theorem 1}\right) .
\end{align*}
So it remains to prove Theorem 1.
We begin with simple lemmas:
Lemma 2. Let $n$ be a positive integer. Let $k$ be a positive integer
such that $2^{k-1}<n\leq2^{k}$. Let $i\in\left[ n\right] ^{\prime}$ be such
that $i\geq2^{k}-n$.
(a) Then, both numbers $i$ and $2^{k}-i-1$ are elements of $\left[
n\right] ^{\prime}$, and at least one of them is $\geq2^{k-1}$.
(b) Let $\sigma$ be a nimble permutation of $\left[ n\right]
^{\prime}$. Then, $\sigma\left( i\right) =2^{k}-i-1$.
Proof of Lemma 2. (a) From $i\in\left[ n\right] ^{\prime}$, we obtain
$i\leq n-1<n\leq2^{k}$, so that $2^{k}-i>0$ and thus $2^{k}-i-1\geq0$. But
$i\geq2^{k}-n$, so that $i+n\geq2^{k}$ and thus $2^{k}-i\leq n$. Thus,
$\underbrace{2^{k}-i}_{\leq n}-1\leq n-1$. Combining this with $2^{k}
-i-1\geq0$, we obtain $2^{k}-i-1\in\left[ 0,n-1\right] =\left[ n\right]
^{\prime}$. Thus, both numbers $i$ and $2^{k}-i-1$ are elements of $\left[
n\right] ^{\prime}$ (since $i\in\left[ n\right] ^{\prime}$ by assumption).
It remains to prove that at least one of them is $\geq2^{k-1}$.
Assume the contrary. Thus, none of these two numbers is $\geq2^{k-1}$. Hence,
they are both $<2^{k-1}$. In other words, $i<2^{k-1}$ and $2^{k}-i-1<2^{k-1}$.
From $i<2^{k-1}$, we obtain $i\leq2^{k-1}-1$ (since $i$ and $2^{k-1}$ are
integers). Hence,
\begin{equation}
2^{k}-1=\underbrace{i}_{\leq2^{k-1}-1}+\underbrace{2^{k}-i-1}_{<2^{k-1}
}<2^{k-1}-1+2^{k-1}=\underbrace{2\cdot2^{k-1}}_{=2^{k}}-1=2^{k}-1.
\end{equation}
This is absurd. This contradiction shows that our assumption was wrong. Hence,
the proof of Lemma 2 (a) is complete.
(b) Lemma 2 (a) shows that both numbers $i$ and $2^{k}-i-1$ are
elements of $\left[ n\right] ^{\prime}$, and at least one of them is
$\geq2^{k-1}$. We are thus in one of the following two cases:
Case 1: We have $i\geq2^{k-1}$.
Case 2: We have $2^{k}-i-1\geq2^{k-1}$.
Let us first consider Case 1. In this case, we have $i\geq2^{k-1}$. But the
permutation $\sigma$ is nimble. Hence, the number $i+\sigma\left( i\right)
+1$ is a power of $2$ (by the definition of "nimble"). In other words,
$i+\sigma\left( i\right) +1=2^{m}$ for some $m\in\mathbb{N}$. Consider this
$m$.
From $2^{m}=\underbrace{i}_{\geq2^{k-1}}+\underbrace{\sigma\left( i\right)
}_{\geq0}+\underbrace{1}_{>0}>2^{k-1}$, we obtain $m>k-1$. Thus, $m\geq k$
(since $m$ and $k$ are integers).
On the other hand, $i\leq n-1$ (since $i\in\left[ n\right] ^{\prime}$) and
$\sigma\left( i\right) \leq n-1$ (since $i\in\left[ n\right] ^{\prime}$).
Hence,
\begin{align*}
2^{m} & =\underbrace{i}_{\leq n-1}+\underbrace{\sigma\left( i\right)
}_{\leq n-1}+1\leq\left( n-1\right) +\left( n-1\right) +1=2\underbrace{n}
_{\leq2^{k}}-1\\
& \leq2\cdot2^{k}-1<2\cdot2^{k}=2^{k+1}.
\end{align*}
Hence, $m<k+1$, so that $m\leq k$ (since $m$ and $k$ are integers). Combining
this with $m\geq k$, we obtain $m=k$. Hence, $2^{m}=2^{k}$, so that
$i+\sigma\left( i\right) +1=2^{m}=2^{k}$ and thus $\sigma\left( i\right)
=2^{k}-i-1$. This proves Lemma 2 (b) in Case 1.
Now, let us consider case 2. In this case, we have $2^{k}-i-1\geq2^{k-1}$. But
we know that $2^{k}-i-1$ is an element of $\left[ n\right] ^{\prime}$.
Hence, there exists some $j\in\left[ n\right] ^{\prime}$ such that
$\sigma\left( j\right) =2^{k}-i-1$ (since $\sigma$ is a permutation of
$\left[ n\right] ^{\prime}$). Consider this $j$. We have $\sigma\left(
j\right) =2^{k}-i-1\geq2^{k-1}$. But the permutation $\sigma$ is nimble.
Hence, the number $j+\sigma\left( j\right) +1$ is a power of $2$ (by the
definition of "nimble"). In other words, $j+\sigma\left( j\right) +1=2^{m}$
for some $m\in\mathbb{N}$. Consider this $m$.
From $2^{m}=\underbrace{j}_{\geq2^{k-1}}+\underbrace{\sigma\left( j\right)
}_{\geq0}+\underbrace{1}_{>0}>2^{k-1}$, we obtain $m>k-1$. Thus, $m\geq k$
(since $m$ and $k$ are integers).
On the other hand, $j\leq n-1$ (since $j\in\left[ n\right] ^{\prime}$) and
$\sigma\left( j\right) \leq n-1$ (since $j\in\left[ n\right] ^{\prime}$).
Hence,
\begin{align*}
2^{m} & =\underbrace{j}_{\leq n-1}+\underbrace{\sigma\left( j\right)
}_{\leq n-1}+1\leq\left( n-1\right) +\left( n-1\right) +1=2\underbrace{n}
_{\leq2^{k}}-1\\
& \leq2\cdot2^{k}-1<2\cdot2^{k}=2^{k+1}.
\end{align*}
Hence, $m<k+1$, so that $m\leq k$ (since $m$ and $k$ are integers). Combining
this with $m\geq k$, we obtain $m=k$. Hence, $2^{m}=2^{k}$, so that
$j+\sigma\left( j\right) +1=2^{m}=2^{k}$ and thus $\sigma\left( j\right)
=2^{k}-j-1$. Comparing this with $\sigma\left( j\right) =2^{k}-i-1$, we find
$2^{k}-i-1=2^{k}-j-1$. Hence, $i=j$. Thus, $\sigma\left( i\right)
=\sigma\left( j\right) =2^{k}-i-1$. This proves Lemma 2 (b) in Case 2.
We have now proven Lemma 2 (b) in both Cases 1 and 2. Thus, Lemma 2
(b) always holds.
Lemma 3. Let $n$ be a positive integer. Let $k$ be a positive integer
such that $2^{k-1}<n\leq2^{k}$. Let $\sigma$ be a nimble permutation of
$\left[ n\right] ^{\prime}$.
(a) We have $2^{k}-n<n$.
(b) The map $\sigma$ preserves the two subsets $\left[ 2^{k}
-n\right] ^{\prime}$ and $\left[ 2^{k}-n,n-1\right] $ of $\left[ n\right]
^{\prime}$ (in other words, it maps each of these two subsets into itself).
(c) Let $\alpha$ be the restriction of $\sigma$ to the subset $\left[
2^{k}-n\right] ^{\prime}$, regarded as a map $\left[ 2^{k}-n\right]
^{\prime}\rightarrow\left[ 2^{k}-n\right] ^{\prime}$. Then, $\alpha$ is a
nimble permutation of $\left[ 2^{k}-n\right] ^{\prime}$.
(d) Let $\beta$ be the restriction of $\sigma$ to the subset $\left[
2^{k}-n,n-1\right] $, regarded as a map $\left[ 2^{k}-n,n-1\right]
\rightarrow\left[ 2^{k}-n,n-1\right] $. Then, $\beta$ is the unique
order-reversing permutation of this subset (i.e., it is strictly decreasing as
a map).
Proof of Lemma 3. (a) We have $2^{k}=2\cdot\underbrace{2^{k-1}}
_{<n}<2n=n+n$, so that $2^{k}-n<n$. This proves Lemma 3 (a).
(b) Let $i\in\left[ 2^{k}-n,n-1\right] $. Hence, $2^{k}-n\leq i\leq
n-1$. Thus, $i\in\left[ n\right] ^{\prime}$ and $i\geq2^{k}-n$. Thus, Lemma
2 (b) shows that
\begin{equation}
\sigma\left( i\right) =2^{k}-\underbrace{i}_{\leq n-1}-1\geq2^{k}-\left(
n-1\right) -1=2^{k}-n.
\end{equation}
Combining this with $\sigma\left( i\right) =2^{k}-\underbrace{i}_{\geq
2^{k}-n}-1\leq2^{k}-\left( 2^{k}-n\right) -1=n-1$, we obtain $\sigma\left(
i\right) \in\left[ 2^{k}-n,n-1\right] $.
Now, forget that we fixed $i$. We thus have shown that $\sigma\left(
i\right) \in\left[ 2^{k}-n,n-1\right] $ for each $i\in\left[
2^{k}-n,n-1\right] $. In other words, the map $\sigma$ preserves the subset
$\left[ 2^{k}-n,n-1\right] $ of $\left[ n\right] ^{\prime}$. Since
$\sigma$ is a permutation of $\left[ n\right] ^{\prime}$, we thus conclude
that $\sigma$ also preserves the complementary subset $\left[ n\right]
^{\prime}\setminus\left[ 2^{k}-n,n-1\right] =\left[ 2^{k}-n\right]
^{\prime}$ of $\left[ n\right] ^{\prime}$. Thus, Lemma 3 (b) is proven.
(c) Lemma 3 (b) shows that the permutation $\sigma$ of $\left[
n\right] ^{\prime}$ preserves the subset $\left[ 2^{k}-n\right] ^{\prime}$
of $\left[ n\right] ^{\prime}$. Hence, it restricts to a permutation of this
subset $\left[ 2^{k}-n\right] ^{\prime}$. Thus, $\alpha$ is a well-defined
permutation of $\left[ 2^{k}-n\right] ^{\prime}$. It remains to show that
$\alpha$ is nimble. But this is clear, because $\alpha$ is a restriction of
the nimble permutation $\sigma$. Thus, Lemma 3 (c) is proven.
(d) Lemma 3 (b) shows that the permutation $\sigma$ of $\left[
n\right] ^{\prime}$ preserves the subset $\left[ 2^{k}-n,n-1\right] $ of
$\left[ n\right] ^{\prime}$. Hence, it restricts to a permutation of this
subset $\left[ 2^{k}-n,n-1\right] $. Thus, $\beta$ is a well-defined
permutation of $\left[ 2^{k}-n,n-1\right] $. Each $i\in\left[
2^{k}-n,n-1\right] $ satisfies
\begin{align*}
\beta\left( i\right) & =\sigma\left( i\right) \qquad\left( \text{since
}\beta\text{ is a restriction of }\sigma\right) \\
& =2^{k}-i-1
\end{align*}
(by Lemma 2 (b), since $i\geq2^{k}-n$); therefore, $\beta$ is the
permutation of $\left[ 2^{k}-n,n-1\right] $ sending each $i$ to $2^{k}-i-1$.
In other words, $\beta$ is the unique order-reversing permutation of this
subset (i.e., it is strictly decreasing as a map). Thus, Lemma 3 (d) is proven.
Definition. Let $A$ and $B$ be two disjoint sets. Let $\alpha$ be a
permutation of $A$. Let $\beta$ be a permutation of $B$. Then, the map
\begin{equation}
A\cup B\rightarrow A\cup B,\qquad x\mapsto
\begin{cases}
\alpha\left( x\right) , & \text{if }x\in A;\\
\beta\left( x\right) , & \text{if }x\in B
\end{cases}
\end{equation}
is a permutation of the set $A\cup B$. This permutation is called the union
of $\alpha$ and $\beta$, and is denoted by $\alpha\cup\beta$.
We are now ready to prove Theorem 1:
Proof of Theorem 1. (a) We shall prove Theorem 1 (a) by strong
induction on $n$:
Fix $n\in\mathbb{N}$. Assume (as the induction hypothesis) that for each
$g\in\mathbb{N}$ satisfying $g<n$, there is a unique nimble permutation
$\sigma_{g}$ of $\left[ g\right] ^{\prime}$. We want to prove that there is
a unique nimble permutation $\sigma_{n}$ of $\left[ n\right] ^{\prime}$.
If $n\leq1$, then this is obvious (because there is a unique permutation of
$\left[ n\right] ^{\prime}$ in this case, and its nimbleness is trivially
verified). Thus, WLOG assume that $n>1$. Thus, there exists a unique positive
integer $k$ satisfying $2^{k-1}<n\leq2^{k}$ (namely, this $k$ is the smallest
positive integer $\ell$ satisfying $2^{\ell}\geq n$). Consider this $k$. We
have $2^{k}-n<n$ (this is proven as in Lemma 3 (a)). Hence, the induction
hypothesis (applied to $g=2^{k}-n$) shows that there is a unique nimble
permutation $\sigma_{2^{k}-n}$ of $\left[ 2^{k}-n\right] ^{\prime}$.
Consider this $\sigma_{2^{k}-n}$.
Let $\gamma$ be the unique order-reversing permutation of the interval
$\left[ 2^{k}-n,n-1\right] $ (that is, the unique strictly decreasing map
$\left[ 2^{k}-n,n-1\right] \rightarrow\left[ 2^{k}-n,n-1\right] $).
Explicitly, $\gamma$ is given by $\gamma\left( i\right) =2^{k}-i-1$ for each
$i\in\left[ 2^{k}-n,n-1\right] $.
The permutations $\sigma_{2^{k}-n}$ and $\gamma$ are permutations of two
complementary subsets of $\left[ n\right] ^{\prime}$ (namely, of the subsets
$\left[ 2^{k}-n\right] ^{\prime}$ and $\left[ 2^{k}-n,n-1\right] $,
respectively). Hence, their union $\sigma_{2^{k}-n}\cup\gamma$ is a
well-defined permutation of $\left[ n\right] ^{\prime}$. Furthermore, this
permutation $\sigma_{2^{k}-n}\cup\gamma$ is nimble.
[Proof. Let $i\in\left[ n\right] ^{\prime}$. We must prove that the number
$i+\left( \sigma_{2^{k}-n}\cup\gamma\right) \left( i\right) +1$ is a power
of $2$.
If $i\in\left[ 2^{k}-n\right] ^{\prime}$, then $\left( \sigma_{2^{k}-n}
\cup\gamma\right) \left( i\right) =\sigma\left( i\right) $ and thus
$i+\underbrace{\sigma\left( i\right) }_{=\sigma_{2^{k}-n}\left( i\right)
}+1=i+\sigma_{2^{k}-n}\left( i\right) +1$ is a power of $2$ (since
$\sigma_{2^{k}-n}$ is nimble). Thus, if $i\in\left[ 2^{k}-n\right] ^{\prime
}$, then we are done. Hence, we WLOG assume that $i\notin\left[
2^{k}-n\right] ^{\prime}$. Hence, $i\in\left[ n\right] ^{\prime}
\setminus\left[ 2^{k}-n\right] ^{\prime}=\left[ 2^{k}-n,n-1\right] $.
Therefore, $\left( \sigma_{2^{k}-n}\cup\gamma\right) \left( i\right)
=\gamma\left( i\right) =2^{k}-i-1$ (by the definition of $\gamma$), so that
$i+\left( \sigma_{2^{k}-n}\cup\gamma\right) \left( i\right) +1=2^{k}$ is a
power of $2$. This completes this proof.]
Thus, there exists at least one nimble permutation of $\left[ n\right]
^{\prime}$ (namely, $\sigma_{2^{k}-n}\cup\gamma$).
Now, let $\sigma$ be a nimble permutation of $\left[ n\right] ^{\prime}$.
Lemma 3 (b) shows that the map $\sigma$ preserves the two subsets $\left[
2^{k}-n\right] ^{\prime}$ and $\left[ 2^{k}-n,n-1\right] $ of $\left[
n\right] ^{\prime}$.
Let $\alpha$ be the restriction of $\sigma$ to the subset $\left[
2^{k}-n\right] ^{\prime}$, regarded as a map $\left[ 2^{k}-n\right]
^{\prime}\rightarrow\left[ 2^{k}-n\right] ^{\prime}$. Lemma 3 (c) shows
that $\alpha$ is a nimble permutation of $\left[ 2^{k}-n\right] ^{\prime}$.
Thus, $\alpha=\sigma_{2^{k}-n}$ (since $\sigma_{2^{k}-n}$ is the unique nimble
permutation $\sigma_{2^{k}-n}$ of $\left[ 2^{k}-n\right] ^{\prime}$).
Let $\beta$ be the restriction of $\sigma$ to the subset $\left[
2^{k}-n,n-1\right] $, regarded as a map $\left[ 2^{k}-n,n-1\right]
\rightarrow\left[ 2^{k}-n,n-1\right] $. Lemma 3 (d) shows that $\beta$
is the unique order-reversing permutation of this subset. Thus, $\beta=\gamma$
(since the unique order-reversing permutation of this subset is $\gamma$).
The permutation $\sigma$ is the union of the permutations $\alpha$ and $\beta$
(since $\alpha$ and $\beta$ are the restrictions of $\sigma$ to two
complementary subsets). In other words, $\sigma=\alpha\cup\beta$. In view of
$\alpha=\sigma_{2^{k}-n}$ and $\beta=\gamma$, this rewrites as $\sigma
=\sigma_{2^{k}-n}\cup\gamma$.
Now, forget that we fixed $\sigma$. We thus have shown that each nimble
permutation $\sigma$ of $\left[ n\right] ^{\prime}$ satisfies $\sigma
=\sigma_{2^{k}-n}\cup\gamma$. Hence, there exists at most one nimble
permutation of $\left[ n\right] ^{\prime}$. Since we already know that there
exists at least one such permutation, we thus conclude that there exists
exactly one such permutation. In other words, there is a unique nimble
permutation $\sigma_{n}$ of $\left[ n\right] ^{\prime}$. This completes the
induction step. Hence, Theorem 1 (a) is proven.
(b) We shall prove Theorem 1 (b) by strong induction on $n$:
Fix $n\in\mathbb{N}$. Assume (as the induction hypothesis) that for each
$g\in\mathbb{N}$ satisfying $g<n$, the unique nimble permutation $\sigma_{g}$
of $\left[ g\right] ^{\prime}$ has sign $\left( -1\right) ^{\sigma_{g}
}=\left( -1\right) ^{g\left( g-1\right) /2}$. We want to prove that the
unique nimble permutation $\sigma_{n}$ of $\left[ n\right] ^{\prime}$ has
sign $\left( -1\right) ^{\sigma_{n}}=\left( -1\right) ^{n\left(
n-1\right) /2}$.
If $n\leq1$, then this is obvious. Hence, WLOG assume that $n>1$. Thus, there
exists a unique positive integer $k$ satisfying $2^{k-1}<n\leq2^{k}$ (namely,
this $k$ is the smallest positive integer $\ell$ satisfying $2^{\ell}\geq n$).
Consider this $k$. Notice that $k$ is positive; thus, $2^{k}$ is even.
We have $2^{k}-n<n$ (this is proven as in Lemma 3 (a)). Hence, the
induction hypothesis (applied to $g=2^{k}-n$) shows that the unique nimble
permutation $\sigma_{2^{k}-n}$ of $\left[ 2^{k}-n\right] ^{\prime}$ has sign
\begin{equation}
\left( -1\right) ^{\sigma_{2^{k}-n}}=\left( -1\right) ^{\left(
2^{k}-n\right) \left( 2^{k}-n-1\right) /2}.
\end{equation}
Let $\gamma$ be the unique order-reversing permutation of the interval
$\left[ 2^{k}-n,n-1\right] $ (that is, the unique strictly decreasing map
$\left[ 2^{k}-n,n-1\right] \rightarrow\left[ 2^{k}-n,n-1\right] $).
Explicitly, $\gamma$ is given by $\gamma\left( i\right) =2^{k}-i-1$ for each
$i\in\left[ 2^{k}-n,n-1\right] $.
A well-known fact says the following: If $m\in\mathbb{N}$, and if $M$ is an
$m$-element set of integers, then the unique order-reversing permutation of
the set $M$ has sign $\left( -1\right) ^{m\left( m-1\right) /2}$. Applying
this to $m=2n-2^{k}$ and $M=\left[ 2^{k}-n,n-1\right] $, we conclude that
the unique order-reversing permutation of the set $\left[ 2^{k}-n,n-1\right]
$ has $\left( -1\right) ^{\left( 2n-2^{k}\right) \left( 2n-2^{k}
-1\right) /2}$. Since this permutation is $\gamma$, we thus have proven that
\begin{equation}
\left( -1\right) ^{\gamma}=\left( -1\right) ^{\left( 2n-2^{k}\right)
\left( 2n-2^{k}-1\right) /2}.
\end{equation}
In the proof of Theorem 1 (a), we have shown that each nimble permutation
$\sigma$ of $\left[ n\right] ^{\prime}$ satisfies $\sigma=\sigma_{2^{k}
-n}\cup\gamma$. Applying this to $\sigma=\sigma_{n}$, we conclude that
$\sigma_{n}$ satisfies $\sigma_{n}=\sigma_{2^{k}-n}\cup\gamma$.
But if $A$ and $B$ are two disjoint finite sets, and if $\alpha$ and $\beta$
are permutations of $A$ and $B$ (respectively), then the union $\alpha
\cup\beta$ of $\alpha$ and $\beta$ has sign $\left( -1\right) ^{\alpha
\cup\beta}=\left( -1\right) ^{\alpha}\left( -1\right) ^{\beta}$. Applying
this to $A=\left[ 2^{k}-n\right] ^{\prime}$, $B=\left[ 2^{k}-n,n-1\right]
$, $\alpha=\sigma_{2^{k}-n}$ and $\beta=\gamma$, we conclude that the union
$\sigma_{2^{k}-n}\cup\gamma$ has sign
\begin{align*}
\left( -1\right) ^{\sigma_{2^{k}-n}\cup\gamma} & =\underbrace{\left(
-1\right) ^{\sigma_{2^{k}-n}}}_{=\left( -1\right) ^{\left( 2^{k}-n\right)
\left( 2^{k}-n-1\right) /2}}\underbrace{\left( -1\right) ^{\gamma}
}_{=\left( -1\right) ^{\left( 2n-2^{k}\right) \left( 2n-2^{k}-1\right)
/2}}\\
& =\left( -1\right) ^{\left( 2^{k}-n\right) \left( 2^{k}-n-1\right)
/2}\left( -1\right) ^{\left( 2n-2^{k}\right) \left( 2n-2^{k}-1\right)
/2}\\
& =\left( -1\right) ^{\left( 2^{k}-n\right) \left( 2^{k}-n-1\right)
/2+\left( 2n-2^{k}\right) \left( 2n-2^{k}-1\right) /2}\\
& =\left( -1\right) ^{n\left( n-1\right) /2}
\end{align*}
(since
\begin{equation}
\left( 2^{k}-n\right) \left( 2^{k}-n-1\right) /2+\left( 2n-2^{k}\right)
\left( 2n-2^{k}-1\right) /2\equiv n\left( n-1\right) /2\operatorname{mod}2
\end{equation}
(because
\begin{align*}
& \left( 2^{k}-n\right) \left( 2^{k}-n-1\right) /2+\left( 2n-2^{k}
\right) \left( 2n-2^{k}-1\right) /2-n\left( n-1\right) /2\\
& =\left( n-2^{k}\right) \underbrace{\left( 2n-2^{k}\right)
}_{\substack{\equiv0\operatorname{mod}2\\\text{(since }2^{k}\text{ is even)}
}}\equiv0\operatorname{mod}2
\end{align*}
)). In view of $\sigma_{n}=\sigma_{2^{k}-n}\cup\gamma$, this rewrites as
$\left( -1\right) ^{\sigma_{n}}=\left( -1\right) ^{n\left( n-1\right)
/2}$. So we have proven that the unique nimble permutation $\sigma_{n}$ of
$\left[ n\right] ^{\prime}$ has sign $\left( -1\right) ^{\sigma_{n}
}=\left( -1\right) ^{n\left( n-1\right) /2}$. This completes the induction
step. Hence, Theorem 1 (b) is proven.
Remark. Our recursive proof of Theorem 1 (a) can be used to show that
the permutation $\sigma_{n}$ is an involution (i.e., it equals its own
inverse) and has no fixed points except possibly $1$ (because the permutation
$\gamma$ is an order-reversing permutation of a set of even size, and such
permutations never have fixed points). Thus, the cycle type of $\sigma_{n}$ is
$\left( \underbrace{2,2,\ldots,2}_{n/2\text{ times}}\right) $ when $n$ is
even, and $\left( \underbrace{2,2,\ldots,2}_{\left(n-1\right)/2\text{ times}},1\right) $
otherwise. This gives another way of computing the sign of $\sigma_{n}$, and
thus of proving Theorem 1 (b).
